Problem 51 - Calculus A New Horizon

Question

Functions of the form $$f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} $$ arise in a wide variety of statistical problems. (a) Use the first derivative test to show that \(f\) has a relative maximum at \(x=0,\) and confirm this by using a graphing utility to graph \(f\) (b) Sketch the graph of $$f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} $$ where \(\mu\) is a constant, and label the coordinates of the relative extrema.

Step-by-Step Solution

Answer

(a) \(f(x)\) has a relative maximum at \(x=0\). (b) For \(f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2}\), the peak is at \(x=\mu\).

Step 1: Differentiate the Function

Find the first derivative of the given function \(f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\) using the product and chain rule.Let \(u(x) = \frac{1}{\sqrt{2 \pi}}\) and \(v(x) = e^{-x^2/2}\). Then, \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Here, \(u'(x) = 0\) and \(v'(x) = e^{-x^2/2} \cdot (-x)\). Thus, \(f'(x) = \frac{1}{\sqrt{2 \pi}} \cdot (-x) e^{-x^2/2} = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2}.\)\.

Step 2: Find Critical Points

Set the first derivative equal to zero to find the critical points:\[-\frac{x}{\sqrt{2\pi}} e^{-x^2/2} = 0.\] This equation is satisfied when \(x = 0\). Thus, \(x = 0\) is a critical point.

Step 3: Apply First Derivative Test

Consider the first derivative test to determine the nature of the critical point \(x = 0\): - For \(x < 0\), \(f'(x) > 0\), thus \(f(x)\) is increasing. - For \(x > 0\), \(f'(x) < 0\), thus \(f(x)\) is decreasing.This shows \(x = 0\) is a relative maximum.

Step 4: Graph Verification Using Utility

Graph the function \(f(x)\) using a graphing utility. Confirm that \(f(x)\) has a peak at \(x = 0\) with symmetric decline, visually verifying the relative maximum at this point.

Step 5: Modify Function for Parameter \(\mu\)

Consider the transform \(f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2}\). This factor shifts the function by \(\mu\) units horizontally. The critical point previously at \(x=0\) is now at \(x=\mu\).

Step 6: Sketch and Label the Graph

Sketch the graph of \(f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2}\), which is similar in shape to the original graph but centered at \(x=\mu\). Label the relative maximum at \( (\mu, \frac{1}{\sqrt{2\pi}}) \), confirming the position of the peak when \(x=\mu\).

Key Concepts

First Derivative Test

The First Derivative Test is a useful method to determine if a critical point is a relative minimum or maximum. In the case of the Gaussian function given by \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} \), we start by calculating the derivative, which involves applying the product and chain rules.
The derivative is \( f'(x) = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} \). To find critical points, we set this first derivative equal to zero, which gives us \( x = 0 \).
After finding \( x = 0 \) as a critical point, the First Derivative Test involves analyzing the sign of the derivative around this point. For \( x < 0 \), the derivative is positive, indicating that \( f(x) \) is increasing. For \( x > 0 \), the derivative is negative, indicating that \( f(x) \) is decreasing. This change in behavior confirms that \( x = 0 \) is a relative maximum.

Graphing Utilities

Graphing utilities can be an incredibly powerful tool for visualizing mathematical functions and their behavior. In this exercise, we use a graphing utility to plot the function \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} \).
The graph should show a symmetric bell-shaped curve, characteristic of Gaussian functions. At \( x = 0 \), you will observe that the curve reaches its peak, showcasing the relative maximum that was previously identified using the First Derivative Test.
Such visualization helps verify analytical findings, providing a more intuitive understanding of the behavior of the function. Additionally, if you adjust the graph to show \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} \), you'll see the curve shifted horizontally by \( \mu \) units, maintaining the same shape but moving the peak along the x-axis.

Relative Extrema

Relative extrema refer to the local highest or lowest points in a function within a specific interval. For the Gaussian function \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} \), the First Derivative Test helped us find a relative extremum at \( x = 0 \).
Since the function changes from increasing to decreasing at this point, \( x = 0 \) is identified as a relative maximum. The height of this maximum can be found by evaluating the function at this point: \( f(0) = \frac{1}{\sqrt{2 \pi}} \).
For generalized Gaussian functions like \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} \), the peak will occur at \( x = \mu \), with the function's maximum value remaining \( \frac{1}{\sqrt{2\pi}} \). Investigating these relative extrema offers insights into the overall shape and key points of the function.

Critical Points

Critical points of a function are values of \( x \) where the function's derivative is zero or undefined, indicating potential maxima or minima. For \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} \), the derivative simplifies to \( f'(x) = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} \).
By setting this derivative to zero, we solve \( -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} = 0 \). The solution is \( x = 0 \), which is a crucial critical point. This is the point where the slope of the graph changes, signifying a shift from increasing to decreasing or vice-versa.
The critical point \( x = 0 \) is thus crucial in determining the relative extremum. When the parameter \( \mu \) is introduced in the generalized function \( f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} \), this critical point shifts to \( x = \mu \), maintaining the function's symmetric nature.